lol, this is a Riemann sum for an integral. The Riemann hypothesis is a completely different thing altogether.
Yes lol, I guess my noobness in maths gave it away haha
P=NP if P=0 or N=1. So as long as neither of those cases apply, P does not equal NP. Do I get $1,000,000 now?
lol like finding both values of the Heisenberg's Uncertainty Principle!::O
The region ofHere's a nice simple one.
Find the locus of:
Note: Try to use a method that minimises algebra.
|z - i| starts from 1 on the imaginary axis.
P(winning on first go) = 1/6
Nice.|z - i| starts from 1 on the imaginary axis.
|z - 1| starts from 1 on the real axis.
Perpendicular bisector of the chord joining those two points is y = x
For |z - i| >= |z - 1|, has to be the region below and including y = x, therefore
lol.. I hope whatever locus you were looking at in your tutorial was more complex than the one you gave.
uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etcP(winning on first go) = 1/6
P(winning on second go) = 5/6 x 5/6 x 1/6
P(winning on third go) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6
By continuing the trend and adding all the possible cases..
P(X winning) = 1/6 + 1/6 x (5/6)^2 + 1/6 x (5/6)^4 + ...
Common ratio of (5/6)^2 therefore it is a geometric progression.
=
I thought that they rolled the dice one at a time.
Because for X to win the second time, he must lose on his roll (probability 5/6), and Y must also lose (probability 5/6).
<a href="http://www.codecogs.com/eqnedit.php?latex=\prod_{k=2}^n\left(1@plus;\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k@plus;1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n@plus;1}{n} =\frac{n@plus;1}{2}\\ ~\\ $For the product to be an integer $n@plus;1$ must be divisible by 2.\\ $\therefore n$ must be odd" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\prod_{k=2}^n\left(1+\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k+1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n+1}{n} =\frac{n+1}{2}\\ ~\\ $For the product to be an integer $n+1$ must be divisible by 2.\\ $\therefore n$ must be odd" title="\prod_{k=2}^n\left(1+\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k+1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n+1}{n} =\frac{n+1}{2}\\ ~\\ $For the product to be an integer $n+1$ must be divisible by 2.\\ $\therefore n$ must be odd" /></a>Oh sorry, I wasn't clear. I was waiting outside for the first year tutorial to finish.
This subject MATH1001 is mostly done by people who did not do Extension 2 Maths, so the idea of finding locus with complex numbers etc is alien to the vast majority.
Anyway new question, a little bit different from what you guys/girls are used to seeing =)
Hint: Use the equivalent of a telescoping sum, but for products.
